Answer:
[tex]x = 16[/tex]
Step-by-step explanation:
Given
[tex]log_{16}(x) + log_4(x) + log_2(x) = 7[/tex]
Required
Solve for x
[tex]log_{16}(x) + log_4(x) + log_2(x) = 7[/tex]
Change base of 16 and base of 4 to base 2
[tex]\frac{log_2(x)}{log_2(16)} + \frac{log_2(x)}{log_2(4)} + log_2(x) = 7[/tex]
Express 16 and 4 as 2^4 and 2^2 respectively
[tex]\frac{log_2(x)}{log_2(2^4)} + \frac{log_2(x)}{log_2(2^2)} + log_2(x) = 7[/tex]
The above can be rewritten as:
[tex]\frac{log_2(x)}{4log_22} + \frac{log_2(x)}{2log_22} + log_2(x) = 7[/tex]
[tex]log_22 = 1[/tex]
So, we have:
[tex]\frac{log_2(x)}{4*1} + \frac{log_2(x)}{2*1} + log_2(x) = 7[/tex]
[tex]\frac{1}{4}log_2(x) + \frac{1}{2}log_2(x) + log_2(x) = 7[/tex]
Multiply through by 4
[tex]4(\frac{1}{4}log_2(x) + \frac{1}{2}log_2(x) + log_2(x)) = 7 * 4[/tex]
[tex]log_2(x) + 2}log_2(x) + 4log_2(x) = 28[/tex]
[tex]7log_2(x) = 28[/tex]
Divide through by 7
[tex]\frac{7log_2(x)}{7} = \frac{28}{7}[/tex]
[tex]log_2(x) = 4[/tex]
Apply the following law of logarithm:
If [tex]log_ab = c[/tex] Then [tex]b = a^c[/tex]
So, we have:
[tex]x = 2^4[/tex]
[tex]x = 16[/tex]
